This application claims the priority of Japanese Patent Application No. 3-120046 filed May 24, 1991 which is incorporated herein by reference.
1. Field of the Invention
The present invention relates to a differential amplifier and to an active filter using the same.
2. Description of the Related Art
Recently, semiconductor integrated circuits have been used for processing voice signals. One type of such circuits uses an active filter to improve the quality of voices. Conventional active filters however employ a discrete structure having capacitors and resistors externally coupled thereto, thus increasing the total number of components of the overall system. To overcome this shortcoming, it is necessary to integrate individual elements constituting an active filter on a semiconductor substrate to incorporate such elements in the substrate.
FIG. 2 illustrates a conventional filter of this type. Resistors R1, R2 and R3 are connected in series, the resistor R3 is further connected to a non-inverting input 21 of an operational amplifier 2. A capacitor C1 is provided between a junction "a" between the resistors R1 and R2 and ground (power supply) GND. A capacitor C2 is provided between the non-inverting input 21 of the operational amplifier 2 and the ground GND. The operational amplifier 2 has an inverting input 22 connected to its output 23, so that the negative feedback of an output signal V.sub.o is effected. A capacitor C3 is provided between the output 23 of the operational amplifier 2 and a junction "b" between the resistors R2 and R3. The positive feedback of the output signal V.sub.o is accomplished by means of the capacitor C3 and the resistor R3. The resistors R2 and R3, the capacitors C2 and C3, and the operational amplifier 2 constitute a basic secondary RC filter. Connecting the resistor R1 and capacitor C1 to this secondary RC filter constitutes a tertiary RC filter. This tertiary RC filter passes the low-frequency component of an analog input signal V.sub.in as an output signal V.sub.out. Moreover, values of various elements shown in FIG. 2 are as follows: R1=R2=R3=10K.OMEGA., C1=6,800 pF, C2=1,000 pF and C3=18,000 pF.
The tertiary RC filter as shown in FIG. 2 has a discrete structure in which the resistors R1 to R3 and the capacitors C1 to C3 are externally coupled. This structure still has the aforementioned problem such that the number of the required system components is increased, enlarging the system.
Proposed as a solution to the above problem is an active filter in which the individual resistors R1 to R3 are replaced with amplifiers 30 as shown in FIG. 7(a), i.e., a so-called biquad filter. In this biquad filter, the individual amplifiers 30 are integrated together with the last-stage operational amplifier 2 on a semiconductor device in order to reduce the number of components and avoid enlargement of the system.
The amplifier 30 shown in FIG. 7(a) will now be described in detail referring to FIG. 1. The amplifier 30 comprises first and second current mirror circuits 3 and 5, and first and second differential amplifier circuits 4 and 6. The first current mirror circuit 3 comprises a pair of npn transistors Tr3 and Tr4. The transistors Tr3 and Tr4 have their collectors connected to a power supply V.sub.cc. A voltage V.sub.1 is applied in common to the bases of the transistors Tr3 and Tr4.
A first differential amplifier circuit 4 comprises a pair of npn transistors Tr5 and Tr6. The first differential amplifier circuit 4 is connected to the first current mirror circuit 3 and also to ground GND via a first constant current supply I1. The transistors Tr5 and Tr6 have collectors respectively connected at output nodes NO4-1 and NO4-2 to the emitters of the transistors Tr3 and Tr4. The emitters of the transistors Tr5 and Tr6 are connected to each other via resistors RE1 and RE2, respectively, and are connected to the first constant current supply I1.
The analog signal V.sub.in, or the output signal V.sub.o from an adjoining amplifier 30, is inputted as an input signal V.sub.in1 to the base of the transistor Tr5. The output signal V.sub.o of the amplifier 30 including the transistor Tr6 is inputted to the base of the transistor Tr6. As a result, the first differential amplifier circuit 4 performs differential amplification on the signals V.sub.in1 and V.sub.o and outputs a complementary signal.
A second current mirror circuit 5 comprises a pair of npn transistors Tr7 and Tr8. The transistors Tr7 and tr8 have emitters connected to ground GND, and bases connected to the collector of the transistor Tr7, respectively.
A second differential amplifier circuit 6 comprises a pair of pnp transistors Tr9 and Tr10. The collectors of the transistors Tr9 and Tr10 are connected to the collectors of the transistors Tr7 and Tr8 of the second current mirror circuit 5. The emitters of the transistor Tr9 and Tr10 are coupled to each other, and are connected to the power supply V.sub.cc via a second constant current supply I2. The base of the pnp transistor Tr9 is connected to the collector of the transistor Tr6. The base of the pnp transistor Tr10 is connected to the collector of the transistor Tr5. The second differential amplifier circuit 6 performs differential amplification on the complementary signal which is outputted from the first differential amplifier circuit 4, and outputs an output signal V.sub.o from the collector of the transistor Tr10.
In the above-structured biquad filter, the amplifier 30 in FIG. 7(a) is equivalent to a circuit shown in FIG. 7(b). This circuit has an operational amplifier 31 of a voltage follower type connected in series to an effective resistor Rx with a large resistance.
Given that various values for physical properties of the biquad filter are expressed as follows, they will have relationships as represented by equations (1) to (8).
______________________________________ Current flowing from the output terminal: I.sub.o Base-emitter voltages of transistors V.sub.BE1, V.sub.BE2 Tr5 and Tr6: Currents flowing across resistors I.sub.R1, I.sub.R2 RE1 and RE2: Thermal voltage: V.sub.T (A voltage based on heat transfer between a base and an emitter of a transistor) Collector currents of transistors I.sub.C1, I.sub.C2 Tr7 and Tr8: Currents of constant current supplies I.sub.R0, I.sub.C0 I1 and I2: ______________________________________
It is assumed that all the resistors have the same resistances, i.e., RE1=RE2=RE. ##EQU1##
Rewriting the equation (3) yields equation (4): EQU (I.sub.R1 -I.sub.R2)/(I.sub.C1 -I.sub.C2)=I.sub.R0 /I.sub.C0 ( 4)
Substituting the equations (2) and (4) into the equation (1) yields equation (5): EQU Rx=RE.multidot.I.sub.R0 /I.sub.C0 ( 5)
Rearranging the equation (5) yields equation (6): EQU I.sub.C0 =RE.multidot.I.sub.R0 /Rx (6)
The cutoff frequency fO of the active filter is represented by the equation (7): EQU f.sub.0 =1/(2.pi..multidot.C1.multidot.Rx) (7)
Rewriting equation (7) yields the equation (8): EQU Rx=1/(2.pi..multidot.C1.multidot.f.sub.0) (8)
Setting the ratio of the capacitors C1 to C2 to C3 shown in FIG. 7 to, for example, EQU C1:C2:C3=1:1/.sqroot.2:.sqroot.2
and setting C1=50 pF yields C2.apprxeq.35 pF and C3.apprxeq.70 pF. Given that the cutoff frequency f.sub.0 is 3 KHz, I.sub.R0 is 50 .mu.A, and RE=10 K.OMEGA., Rx=1.06 M.OMEGA. from the equation (8), and I.sub.C0 .apprxeq.472 nA from the equation (6).
The use of the conventional circuit for processing high frequency signals may be acceptable. However, the use of the conventional circuit with voice frequency bands having a cutoff frequency f.sub.0 =3 KHz, will give rise to the following significant concerns.
Generally, forming a capacitor having a capacitance of, for example, 1 pF on a semiconductor substrate requires an area of 100 .mu..quadrature.. If one tries to integrate the capacitors C1, C2 and C3 respectively having the aforementioned large capacitances of 50 pF, 35 pF and 70 pF on the semiconductor substrate, a large area will be required, lowering the integration density.
To avoid such a reduction in integration density, each of the capacitors C1 to C3 may be designed to have a tenth of the mentioned value. In this case, however, it is apparent from the equation (8) that the value of the effective resistance Rx must be increased by a factor of f.sub.0. From the equation (6), the current value I.sub.C0 of the constant current supply I2 must be reduced to one tenth. Diminishing the minute current I.sub.C0 (.apprxeq.472 nA) of the constant current supply I2 to a tenth however requires very high precision and is practically impossible.